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      <a href="/2018/10/14/groovy-unique/" title="Groovy Unique 方法性能测试" itemprop="url">Groovy Unique 方法性能测试</a>
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		<a href="/about" title="Luo Yuan" target="_blank" itemprop="author">Luo Yuan</a>
		
  <p class="article-time">
    <time datetime="2018-10-14T02:53:00.000Z" itemprop="datePublished"> Published 2018-10-14</time>
    
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			<strong class="toc-title">Contents</strong>
		
			<ol class="toc"><li class="toc-item toc-level-1"><a class="toc-link" href="#%E6%95%B0%E5%AD%97%E5%88%97%E8%A1%A8%E7%9A%84%E5%8E%BB%E9%87%8D"><span class="toc-number">1.</span> <span class="toc-text">数字列表的去重</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#%E6%B5%8B%E8%AF%95%E4%BB%A3%E7%A0%81"><span class="toc-number">1.1.</span> <span class="toc-text">测试代码</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#%E6%B5%8B%E8%AF%95%E7%BB%93%E6%9E%9C"><span class="toc-number">1.2.</span> <span class="toc-text">测试结果</span></a></li></ol></li><li class="toc-item toc-level-1"><a class="toc-link" href="#%E5%AD%97%E7%AC%A6%E4%B8%B2%E5%88%97%E8%A1%A8%E7%9A%84%E5%8E%BB%E9%87%8D"><span class="toc-number">2.</span> <span class="toc-text">字符串列表的去重</span></a><ol class="toc-child"><li class="toc-item toc-level-2"><a class="toc-link" href="#%E6%B5%8B%E8%AF%95%E4%BB%A3%E7%A0%81-1"><span class="toc-number">2.1.</span> <span class="toc-text">测试代码</span></a></li><li class="toc-item toc-level-2"><a class="toc-link" href="#%E6%B5%8B%E8%AF%95%E7%BB%93%E6%9E%9C-1"><span class="toc-number">2.2.</span> <span class="toc-text">测试结果</span></a></li></ol></li><li class="toc-item toc-level-1"><a class="toc-link" href="#Groovy-unique%E6%96%B9%E6%B3%95%E6%BA%90%E7%A0%81%E5%88%86%E6%9E%90"><span class="toc-number">3.</span> <span class="toc-text">Groovy unique方法源码分析</span></a></li><li class="toc-item toc-level-1"><a class="toc-link" href="#%E7%BB%93%E8%AE%BA"><span class="toc-number">4.</span> <span class="toc-text">结论</span></a></li></ol>
		
		</div>
		
		<p>最近系统遇到一个问题就是一个调用一个groovy方法总是长时间卡住，研究发现这个方法中使用了unique方法。身为一个java程序员一直比较习惯用HashSet去过滤重复的数据，看到groovy中有这么一个去重复的方法，觉得挺方便，所以就大咧咧的使用了。但是测试以后发现，这个方法性能真的是很差劲的…… <span id="more"></span></p>
<h1 id="数字列表的去重"><a href="#数字列表的去重" class="headerlink" title="数字列表的去重"></a>数字列表的去重</h1><h2 id="测试代码"><a href="#测试代码" class="headerlink" title="测试代码"></a>测试代码</h2><p>首先是测试数值的比较，我以下代码分别测速list的大小为100， 10000， 1000000。</p>
<pre><code>random = new Random();
performanceReport(100)
performanceReport(10000)
performanceReport(10000000)
def performanceReport(int size) &#123;
    println(&quot;Report for list size $size&quot;)
    def list = [];
    (1..size).each &#123;
        list.add(random.nextInt(1000))
    &#125;
    def start = System.currentTimeMillis();
    list.unique();
    def end = System.currentTimeMillis();
    def dur = end - start;
    dur /= 1000;
    println(&quot;unique use time $dur s&quot;);
    list.clear();
    (1..size).each &#123;
        list.add(random.nextInt(1000))
    &#125;
    start = System.currentTimeMillis();
    def uSet = new HashSet(list.size());
    for (e in list) &#123;
        uSet.add(e);
    &#125;
    end = System.currentTimeMillis();
    dur = end - start;
    dur /= 1000;
    println(&quot;hash set use time $dur s&quot;);
    println(&#39;---------------------&#39;);
&#125;
</code></pre>
<h2 id="测试结果"><a href="#测试结果" class="headerlink" title="测试结果"></a>测试结果</h2><pre><code>Report for list size 100
unique use time 0.006 s
hash set use time 0.006 s
---------------------
Report for list size 10000
unique use time 0.127 s
hash set use time 0.004 s
---------------------
Report for list size 10000000
unique use time 49.479 s
hash set use time 0.279 s
</code></pre>
<p>我重复运行了代码三遍，出来的结果都差不多，所以对比结果基本是确定的：</p>
<ol>
<li>List比较少元素的情况下，比如100以内，运行效率差不了多少</li>
<li>当List元素较多的时候，差异非常明显，特别是大容量List的运行结果，使用unique就绝对是一个错误的选择！</li>
</ol>
<h1 id="字符串列表的去重"><a href="#字符串列表的去重" class="headerlink" title="字符串列表的去重"></a>字符串列表的去重</h1><h2 id="测试代码-1"><a href="#测试代码-1" class="headerlink" title="测试代码"></a>测试代码</h2><p>我使用随机生成的十位英文字符串（只包含a、b、c）进行测试比较。测试运行比较列表大小为100、1000、100000的结果。</p>
<pre><code>random = new Random();
strings = [&#39;a&#39;, &#39;b&#39;, &#39;c&#39;];
performanceReport(100)
performanceReport(1000)
performanceReport(100000)
//生成随机字符串
def genString()&#123;
    def s = &#39;&#39;;
    (1..4).each&#123;
        def index = random.nextInt(strings.size());
        s += strings[index];
    &#125;
    return s;
&#125;
def performanceReport(int size) &#123;
    println(&quot;Report for list size $size&quot;)
    def list = [];
    (1..size).each &#123;
        list.add(genString())
    ｝
    def start = System.currentTimeMillis();
    list.unique();
    def end = System.currentTimeMillis();
    def dur = end - start;
    dur /= 1000;
    println(&quot;unique use time $dur s&quot;);
    list.clear();
    (1..size).each &#123;
        list.add(genString())
    &#125;
    start = System.currentTimeMillis();
    def uSet = new HashSet(list.size());
    for (e in list) &#123;
        uSet.add(e);
    &#125;
    end = System.currentTimeMillis();
    dur = end - start;
    dur /= 1000;
    println(&quot;hash set use time $dur s&quot;);
    println(&#39;---------------------&#39;);
&#125;
</code></pre>
<h2 id="测试结果-1"><a href="#测试结果-1" class="headerlink" title="测试结果"></a>测试结果</h2><p>重复运行代码三遍，发现结果对比的差值都比较固定，所以可以取其中一个结果作为参考。</p>
<pre><code>Report for list size 100
unique use time 0.002 s
hash set use time 0.005 s
---------------------
Report for list size 1000
unique use time 0.049 s
hash set use time 0.001 s
---------------------
Report for list size 100000
unique use time 30.716 s
hash set use time 0.019 s
</code></pre>
<ol>
<li>列表容量较少的时候（100左右），groovy的unique方法性能还是有一点点优势的</li>
<li>大容量的列表去重的时候，依旧比使用HashSet的速度慢了很多。所以如果大列表处理的时候，还是得选用HashSet。</li>
</ol>
<h1 id="Groovy-unique方法源码分析"><a href="#Groovy-unique方法源码分析" class="headerlink" title="Groovy unique方法源码分析"></a>Groovy unique方法源码分析</h1><p>unique方法的源码如下：</p>
<pre><code>public static &lt;T&gt; Collection&lt;T&gt; unique(Collection&lt;T&gt; self, boolean mutate) &#123;
    ArrayList answer = new ArrayList();
    Iterator i$ = self.iterator();
    //双重循环
    while(i$.hasNext()) &#123;
        Object t = i$.next();
        boolean duplicated = false;
        Iterator i$1 = answer.iterator();
        //循环传进来的带重复元素的列表而不是循环已经去重的列表
        while(i$1.hasNext()) &#123;
            Object t2 = i$1.next();
            if(coercedEquals(t, t2)) &#123;
                duplicated = true;
                break;
            &#125;
        &#125;

        if(!duplicated) &#123;
            answer.add(t);
        &#125;
    &#125;

    if(mutate) &#123;
        self.clear();
        self.addAll(answer);
    &#125;

    return (Collection)(mutate?self:answer);
&#125;
</code></pre>
<p>可以发现，非常不幸的两个导致性能底下的代码实现</p>
<ol>
<li>双重循环</li>
<li>在判断重复元素的时候，是去循环原始的List，而不是新构建的容量较小的List，这个也会增加代码执行时间。</li>
</ol>
<h1 id="结论"><a href="#结论" class="headerlink" title="结论"></a>结论</h1><p>一直以为这种框架性质的代码，一定是千锤百炼，写得优雅又高效的，不过事实就是：代码写得好看，性能并不一定好！<br>Groovy方法虽然便捷，但是使用内置的方法还是得谨慎，毕竟要处理超过100的列表是经常的事。</p>
  
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